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Let be $B$ an operator and $\left|\Psi\right>$, $\left|\Phi\right>$ two states (not necessarily equals).

What is the meaning of a matrix element $\left<\Psi\right| B \left|\Phi\right>\neq 0$ for $\left|\Psi\right> \neq \left|\Phi\right>$ ? Can i think that this matrix element isn't zero due to a perturbation that doesn't allow to diagonalize the operator $B$ ?

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closed as unclear what you're asking by Emilio Pisanty, DavePhD, Kyle Kanos, Kyle Oman, Jim Jun 10 '14 at 15:38

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Not quite.

The matrix elements you are talking about are called "off-diagonal" for obvious reasons: If you'd write down the matrix, these elements would occur somewhere other than on the diagonal.

A non-zero off-diagonal element of an operator $B$ does not necessarily mean that you cannot diagonalize $B$ at all. It just means that in the currently used basis, $B$ is not diagonal, simple as that.

You can make some connections to perturbation theory, though. Let's say we have a "simple" Hamiltonian $H_0$ and a perturbation $V$ so that the full Hamiltonian is $H_0 + V$. Then usually that means that we know the eigenstates of $H_0$ and in the basis made up of those eigenstates, $H_0$ will be diagonal.

The perturbation, however, isn't necessarily diagonal in that eigenbasis. If it was, then $H_0 + V$ would be as easy to diagonalize as $H_0$. Now, what does a non-zero matrix element of $V$ in the eigenbasis of $H_0$ mean? It means that the perturbation "mixes" those two states. It means that the true eigenstates of $H_0 + V$ will contain states that have some probability to be in $\Psi$ and some probability to be in $\Phi$.

In the context of time-dependent perturbation theory, it means that if you start in state $\Phi$, then in the unperturbed system you'd stay in $\Phi$, but due to the perturbation there's a probability that the system transitions to state $\Psi$, given by Fermi's golden rule.

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